Painted Inner
Painted Inner
A spherical ball of radius 1 is covered with ink and placed in the hollow area between two concentric s?
A spherical ball radius 1 is covered with ink and placed in the hollow area between two concentric spheres with radii of 3 and 5. The ball rolled signed in this region, still not touching the surface inside of the outer sphere and the outer surface of the inner sphere. Suppose that a region of zone 1 in the outer sphere is painted with ink on the scroll wheel. Find the area near the inner sphere, which is inked by the trackball.
I think this is supposed to be on the inverse square law. Unfortunately, this hypothetical problem seems a bit poorly drafted. For example, no mention was made about whether the concentric spheres can be run separately from each other. If they could rotate separately then the area could be the same, assuming there is friction and time never touched the ball signed the same point in each area over a time. The reason is that this is the outermost of turnovers signed in the same proportion as the deepest (in the outermost regions referred to the center depths of concentric spheres). If these areas could rotate, could sort of imagine how gears and realize that both the external and Internal would move the same amount. Nitpick could in this scenario, but nevermind that I believe this development will .... Imagine signed the ball made a complete revolution in these concentric spheres (assuming concentric spheres can not rotate independently of one another and the ink makes the ink frictionless ball and still concentric spheres mark). After a complete rotation, the outer sphere was signed in a circle of radius 5, while the private sphere has been signed within a radius of 3. Next Imagine Imagine a ball signed instead inked equivalent to a square in the outer sphere. Suppose that the sides are of length 1, then we know that of the most intimate sphere is like a square with a side of 3 / 5, as is the ratio of their radii. Areas of a square are the times of self-SIDE Thus, when the place is the outermost signed a patch of 1 square unit, then the closest would be a corresponding patch (3 / 5) * (3 / 5) = 9 / 25 square units. Hope this helps!
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